Hints for Sample Final problem #2 --------------------------------- Problem #2 on the sample final is a tricky problem, but you can do it in at least three seemingly different ways. (Maybe this is why it's so tricky!) (A) You can do it using the directional derivative. This is probably the method most people are trying, but there's a tricky step at the beginning. (B) You can do it using the chain rule, but there's a tricky step to set it up this way. (C) Once you've figured out the tricky step for (B), you can actually solve this problem using single variable calculus! Here are a few comments on each approach. METHOD A -------- Using the given information, you should be able to find the directional derivative of T(x,y,z) at the point (2,3,1) in the direction of the vector which goes from (2,3,1) to (3,4,3). BE CAREFUL - you're not done. This gives you the (instantaneous) change in temperature per change in kilometer in that direction. You're supposed to give an answer in terms of change in temperature per second. So you need to figure out how to use the rest of the given information to get your final answer. METHOD B -------- You're trying to find a partial derivative of T, which is a function of x, y, and z. But these three variables are themselves functions of t (time). This sounds like the chain rule! [Read the discussion from the bottom of page 203 through 204, but realize that the notation is a little deceiving; whenever they say "du/dx" for example, they really mean "du/dx evaluated at (x(s,t),y(s,t))."] To use the chain rule in our case, you need to do two things: (1) Find a parametrization f(t)=(x(t),y(t),z(t)) for the path -- i.e. the line -- which goes straight from (2,3,1) to (3,4,3) at a constant speed of 5. (Remember, the "speed" is the length of the tangent vector f'(t).) This is actually pretty tricky. (2) Set up the chain rule, where you're interested in T(f(t)), so the derivative is JT(f(t))Jf(t). You're going to want to evaluate this at whatever value of t gives you f(t)=(2,3,1). Note that T has three inputs and one output, while f has one input and three outputs, so the chain rule looks like this: [dT/dx dT/dy dT/dz ] [ dx/dt ] [ dy/dt ] [ dz/dt ] Which multiplies out to: dT dT dx dT dy dT dz -- = -- -- + -- -- + -- -- dt dx dt dy dt dz dt Remember, the partial derivatives like dT/dx here need to be evaluated at f(t) here. METHOD C -------- Suppose you've found the parametrization f(t) mentioned above. Let's say f(0)=(2,3,1), although you could have it set up a little differently. f(t) tells us what x, y, and z are in terms of t, so you could plug that into T(x,y,z) and get a new function Temp(t) which expresses temperature in terms of time. Then this is a single variable calculus problem, and you're looking for Temp'(0), the derivative at 0. --------------- This sounds like a complicated problem, but in fact you can do it using method A in a few minutes once you realize the trick at the end. I've included the other approaches here since a few people have tried them. The answer, incidentally, is 205/Sqrt[6]. Hope this helps! Jonathan Rogness (rogness@math.umn.edu)